Question

Reduce the following PDE to a set of three ODEs by the method of separation of
variables (1/r)(∂/∂r)(r(∂V/∂r)+(1/r^2)(∂^2V/∂θ^2)+(∂^2V/∂z^2)+(k^2)V=0

Answer 1

Answer on Question #76479 – Math – Differential Equations

Question

Reduce the following PDE to a set of three ODEs by the method of separation of variables

$$\frac {1}{r} \frac {\partial}{\partial r} \left(r \left(\frac {\partial V}{\partial r}\right)\right) + \left(\frac {1}{r ^ {2}}\right) \left(\frac {\partial^ {2} V}{\partial \theta^ {2}}\right) + \left(\frac {\partial^ {2} V}{\partial z ^ {2}}\right) + (k ^ {2}) V = 0$$

Solution

Assume

$$V (r, \theta , z) = R (r) \Theta (\theta) Z (z)$$ $$\Theta Z \left(\frac {1}{r}\right) \frac {d}{d r} \left(r \left(\frac {d R}{d r}\right)\right) + R Z \left(\frac {1}{r ^ {2}}\right) \left(\frac {d ^ {2} \Theta}{d \theta^ {2}}\right) + R \Theta \left(\frac {d ^ {2} Z}{d z ^ {2}}\right) + (k ^ {2}) R \Theta Z = 0$$

Divide through by $V$ :

$$\frac {1}{R} \left(\frac {1}{r}\right) \frac {d}{d r} \left(r \left(\frac {d R}{d r}\right)\right) + \frac {1}{\Theta} \left(\frac {1}{r ^ {2}}\right) \left(\frac {d ^ {2} \Theta}{d \theta^ {2}}\right) + \frac {1}{Z} \left(\frac {d ^ {2} Z}{d z ^ {2}}\right) + k ^ {2} = 0$$ $$\frac {1}{R} \left(\frac {1}{r}\right) \frac {d R}{d r} + \frac {1}{R} \frac {d ^ {2} R}{d r ^ {2}} + \frac {1}{\Theta} \left(\frac {1}{r ^ {2}}\right) \left(\frac {d ^ {2} \Theta}{d \theta^ {2}}\right) + k ^ {2} = - \frac {1}{Z} \left(\frac {\partial^ {2} Z}{\partial z ^ {2}}\right)$$

In the above equation the left-hand side depends on $r$ and $\theta$ , while the right-hand side depends on $z$ . The only way these two members are going to be equal for all values of $r, \theta$ and $z$ is when both of them are equal to a constant. Let us define such a constant as $-l^2$ .

With this choice for the constant, we obtain:

$$\frac {d ^ {2} Z}{d z ^ {2}} - l ^ {2} Z = 0$$

The general solution of this equation is:

$$Z (z) = a _ {1} e ^ {l z} + a _ {2} e ^ {- l z}$$

Such a solution, when considering the specific boundary conditions, will allow $Z(z)$ to go to zero for $z$ going to $\pm \infty$ , which makes physical sense. If we had given the constant a value of $l^2$ , we would have had periodic trigonometric functions, which do not tend to zero for $z$ going to infinity.

Once sorted the $z$ -dependency, we need to take care of $r$ and $\theta$ .

$$\frac {1}{R} \left(\frac {1}{r}\right) \frac {d R}{d r} + \frac {1}{R} \frac {d ^ {2} R}{d r ^ {2}} + \frac {1}{\Theta} \left(\frac {1}{r ^ {2}}\right) \left(\frac {d ^ {2} \Theta}{d \theta^ {2}}\right) = - (k ^ {2} + l ^ {2})$$ $$\frac {r}{R} \frac {d R}{d r} + \frac {r ^ {2}}{R} \frac {d ^ {2} R}{d r ^ {2}} + (k ^ {2} + l ^ {2}) r ^ {2} = - \frac {1}{\Theta} \left(\frac {d ^ {2} \Theta}{d \theta^ {2}}\right)$$

Again we are in a situation where the only way a solution can be found for the above equation is when both members are equal to a constant. This time we select a positive constant, which we call $m^2$ . The equation for $\Theta$ becomes, then:

$$\frac {d ^ {2} \theta}{d \theta^ {2}} + m ^ {2} \theta = 0$$

Its general solution can be written as:

$$\theta (\theta) = b _ {1} \sin (m \theta) + b _ {2} \cos (m \theta)$$

This solution is well suited to describe the variation for an angular coordinate like $\theta$ . Had we chosen to set both members of equation equal to a negative number, we would have ended up with exponential functions with a different value assigned to $\theta(\theta)$ for each 360 degrees turn, a clear non-physical solution.

Last to be examined is the r-dependency. We have:

$$\frac {r}{R} \frac {d R}{d r} + \frac {r ^ {2}}{R} \frac {d ^ {2} R}{d r ^ {2}} + (k ^ {2} + l ^ {2}) r ^ {2} = m ^ {2}$$ $$r ^ {2} \frac {d ^ {2} R}{d r ^ {2}} + r \frac {d R}{d r} + \left((k ^ {2} + l ^ {2}) r ^ {2} - m ^ {2}\right) R = 0$$

The equation $(\ast)$ is a well-known equation of mathematical physics called parametric Bessel's equation. With a simple linear transformation of variable, $x = (\sqrt{k^2 + l^2})r$ , equation $(\ast)$ is readily changed into a Bessel's equation:

$$\frac {d R}{d r} = \frac {d R}{d x} \frac {d x}{d r} = \left(\sqrt {k ^ {2} + l ^ {2}}\right) R ^ {\prime}$$ $$\frac {d ^ {2} R}{d r ^ {2}} = \frac {d}{d x} \left(\left(\sqrt {k ^ {2} + l ^ {2}}\right) R ^ {\prime}\right) \frac {d x}{d r} = (k ^ {2} + l ^ {2}) R ^ {\prime \prime}$$ $$\frac {x ^ {2}}{k ^ {2} + l ^ {2}} (k ^ {2} + l ^ {2}) R ^ {\prime \prime} + \frac {x}{\sqrt {k ^ {2} + l ^ {2}}} \left(\sqrt {k ^ {2} + l ^ {2}}\right) R ^ {\prime} + (x ^ {2} - m ^ {2}) R = 0$$ $$x ^ {2} R ^ {\prime \prime} + x R ^ {\prime} + (x ^ {2} - m ^ {2}) R = 0$$

where $R''$ and $R'$ indicate first and second derivatives with respect to $x$ .

In what follows we will assume that $m$ is a real, non-negative number.

Linearly independent solutions are typically denoted $J_{m}(x)$ (Bessel Functions) and $N_{m}(x)$ (Neumann Functions).

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